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Popular >

y^'=y(1+x^2)

Solution

y^{^{'}} = y (1 + x^{2})

Solution

y = e^{x + \frac{x ^{3}}{3} + c_{1}}

Solution steps

y^{'} (x) = y (1 + x^{2})

Solve separable ODE:

y = e^{x + \frac{x ^{3}}{3} + c_{1}}

y = e^{x + \frac{x ^{3}}{3} + c_{1}}

Popular Examples

e^xcos(y)dx-xe^xsin(y)dy=0

e^{x} cos (y) d x - x e^{x} sin (y) d y = 0

y^'+9y=0,y(0)=2

y^{^{'}} + 9 y = 0, y (0) = 2

(x^2+7)y^'=x(y^2+9)

(x^{2} + 7) y^{^{'}} = x (y^{2} + 9)

y^'=(2x)/(y+x^2y),y(0)=-2

y^{^{'}} = \frac{2 x}{y + x ^{2} y}, y (0) = - 2

(dy)/(dx)=55yx^{10}

\frac{d y}{d x} = 55 y x^{10}

Frequently Asked Questions (FAQ)

What is the solution for y^'=y(1+x^2) ?
The solution for y^'=y(1+x^2) is y=e^{x+(x^3)/3+c_{1}}